Two electric lamps rated at 60\,\text{W}, 220\,\text{V} and 100\,\text{W}, 220\,\text{V} r… — JAMB UTME

Two electric lamps rated at $60\,\text{W}, 220\,\text{V}$ and $100\,\text{W}, 220\,\text{V}$ respectively are connected in series across a $220\,\text{V}$ supply line. What is the total electric power dissipated by the combination?

$37.5\,\text{W}$ Answer
B
$80.0\,\text{W}$
C
$160.0\,\text{W}$
D
$40.0\,\text{W}$

Answer

The total electric power dissipated by the two lamps in series is

37.5\,\text{W}

For two appliances designed for the same rated voltage

V

connected in series across that voltage

V

, the equivalent power is given by the formula

P_{\text{total}} = \frac{P_1 P_2}{P_1 + P_2}

. Substituting

P_1 = 60\,\text{W}

and

P_2 = 100\,\text{W}

gives

P_{\text{total}} = \frac{60 \times 100}{60 + 100} = 37.5\,\text{W}

Step-by-Step Solution

Calculate the resistance of each lamp from its rating.

R_1 = \frac{V^2}{P_1} = \frac{220^2}{60} = 806.67\,\Omega

and

R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484.00\,\Omega

Electrical devices are rated at a specific voltage, allowing their resistance to be determined via

R = \frac{V^2}{P}

Find the total resistance of the series circuit.

R_{\text{total}} = R_1 + R_2 = 806.67 + 484.00 = 1290.67\,\Omega

Resistors in series add directly.

Calculate total power drawn from the

220\,\text{V}

supply.

P_{\text{total}} = \frac{V^2}{R_{\text{total}}} = \frac{220^2}{1290.67} = 37.5\,\text{W}

Total power is total voltage squared divided by equivalent resistance.

Key Concept

Power combination in series circuits

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic